0
$\begingroup$

Suppose the sequence $\{x_i\}$ does not converge to $y$. I want to prove that then there exists a neighborhood of $U$ and a subsequence of $\{x_i\}$ that lies completely outside $U$.

In my attempt, I prove the contrapositive instead:

For any neighborhood $U$ around $y$ there exists no subsequence of $\{x_i\}$ that lies completely outside $U$. Prove that $\{x_i\}\rightarrow y$.

Since $\{x_i\}$ has no subsequence completely outside $U$, it has no elements outside $U$ (otherwise we could simply choose a subsequence consisting of 1 element that would be outside $U$). That means $\{x_i\}$ lies within any neighborhood of $U$. If we now define a neighborhood in the form of an open ball $S_r(y)$ and let $r\rightarrow 0$ all elements of $\{x_i\}$ approach $y$, thus $\{x_i\}\rightarrow y$.

Is this proof correct?

  • 0
    In general, you cannot pick "a subsequence consisting of 1 element": given a sequence $\{x_i\}$ (which, remember, is defined to be *a function* $x:\Bbb N\to X$, and $x_n$ is a shorthand notation for $x(n)$), a subsequence is a function $y:\Bbb N\to X$ which can be written as $y=x\circ f$ for some *strictly increasing* function $f:\Bbb N\to\Bbb N$. – 2017-02-22
  • 0
    Okay! How about this? Either $\{x_i\}$ has infinitely many elements outside $U$ (in which case there would exist a subsequence completely outside $U$) or it has finitely many elements outside $U$. If it has finitely many then we can look at the convergence of those elements that lie within $U$, and realize that the sequence would converge to $y$. – 2017-02-22

0 Answers 0